Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{0}^{1} \frac{\ln x}{x} d x\)

Short answer

The integral evaluates to -1.

Step-by-step solution

Analyze the Behavior at the Lower Limit

The integral \( \int_{0}^{1} \frac{\ln x}{x} \, dx \) has a problem at the lower limit \( x = 0 \), where \( \ln x \) diverges to \(-\infty\). Thus, the integral might be improper at this limit.

Set Up the Limit Definition for the Improper Integral

To properly evaluate the improper integral, we use a limit: \[ \int_{0}^{1} \frac{\ln x}{x} \, dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{\ln x}{x} \, dx.\] This effectively avoids dividing by zero directly and properly handles the singularity at \( x = 0 \).

Apply Integration by Parts

To solve \( \int \frac{\ln x}{x} \, dx \), use integration by parts. Choose \( u = \ln x \) and \( dv = \frac{1}{x} \, dx \), then \( du = \frac{1}{x} \, dx \) and \( v = x \). The integration by parts formula \( \int u \, dv = uv - \int v \, du \) gives us: \[ \int \frac{\ln x}{x} \, dx = \ln x \cdot x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C. \]

Evaluate the Definite Integral with Limits

Substitute the result from integration by parts into the limit: \[ \lim_{a \to 0^+} \left[ x \ln x - x \right]_a^1 = \lim_{a \to 0^+} \left[ (1 \cdot \ln 1 - 1) - (a \ln a - a) \right]. \] Evaluate at \( x = 1 \): \( 1 \ln 1 - 1 = -1 \). Evaluate the limit as \( x \to a \): \( \lim_{a \to 0^+} (a \ln a - a) \).

Determine the Limit for the Improper Part

Investigate \( \lim_{a \to 0^+} (a \ln a - a) \): \( a \ln a \approx 0 \cdot (-\infty) \), leading to an indeterminate form \( \lim_{a \to 0^+} \frac{\ln a}{1/a} \). Applying L'Hôpital's rule twice, \( \lim_{a \to 0^+} \ln a / (1/a) = \lim_{a \to 0^+} -a = 0\). Thus, \( \lim_{a \to 0^+} (a \ln a - a) = 0. \) "},{

Determine the Value of the Integral

Combine results: \[ \lim_{a \to 0^+} \left[ -1 - (0) \right] = -1. \] Therefore, the entire integral evaluates to \(-1\).

Key concepts

Integration by Parts

When facing integrals like \( \int \frac{\ln x}{x} \, dx \), integration by parts can come to our rescue. This technique transforms a complex integral into simpler ones by splitting the integrand into two parts. We choose a function to differentiate and another to integrate. In this example:

• Let \( u = \ln x \), so \( du = \frac{1}{x} \, dx \).
• Let \( dv = \frac{1}{x} \, dx \), leading to \( v = x \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we connect the two parts:\[ \int \frac{\ln x}{x} \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C. \]This step simplifies our original integral, offering an expression that forms the foundation for the final evaluation of the definite integral in the problem.

L'Hôpital's Rule

In some scenarios, like evaluating limits that result in indeterminate forms, L'Hôpital's Rule becomes invaluable. This rule helps us tackle limits like \( \lim_{a \to 0^+} (a \ln a - a) \). The challenge here lies in encountering the form \( 0 \cdot (-\infty) \), which is indeterminate.To apply L'Hôpital's Rule effectively:

• Rewrite the limit as a quotient: \( \lim_{a \to 0^+} \frac{\ln a}{1/a} \).
• Identify that both the numerator and denominator approach infinity or zero, an indeterminate form \( \frac{-\infty}{\infty} \).

Apply L'Hôpital's Rule, which requires us to differentiate both the numerator and the denominator, yielding:\[ \lim_{a \to 0^+} \frac{1/a}{-1/a^2} = \lim_{a \to 0^+} -a = 0. \]Thus, L'Hôpital's Rule reveals that the limit simplifies to 0, a crucial step to properly evaluating the integral.

Limit Definition of Integrals

When dealing with improper integrals, it's essential to redefine them using limits. An improper integral like \( \int_{0}^{1} \frac{\ln x}{x} \, dx \) poses challenges due to the behavior of \( \ln x \) at \( x = 0 \). Instead of directly evaluating it, we redefine this as a limit:\[ \int_{0}^{1} \frac{\ln x}{x} \, dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{\ln x}{x} \, dx. \]This redefinition effectively circumvents the singularity at the boundary:

• By introducing parameter \( a \), ranging from some small positive value towards zero, we manage the undefined behavior.
• This limit approach ensures that our integral evaluation respects mathematical rigor and precision.

Incorporating limits helps us bridge the gaps left by indeterminate behavior, paving the way for applying subsequent steps like integration by parts and L'Hôpital's Rule to evaluate within finite boundaries. Ultimately, this leads us to a well-defined value for the integral, such as the solution found, \(-1\).